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Block held in equilibrium on incline by Force F, Find magnitidue of Force F?

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A 5.79KG block is held in equilibrium on an incline by the horizontal force F, as shown in the figure. Determine the magnitude of F. Answer in units N. The figure is in the attachment, problem 008.


    2. Relevant equations
    SigmaF=0
    Fgx=mg(SinTheta)
    Ffr(force of friction)= mu(fn)
    Fn= Fgy
    Fgy= mg(CosTheta)

    3. The attempt at a solution
    Through those equations, I eventually got this:
    F= mg(SinTheta) - mu[mg(cosTheta)]
    F= 6.052

    That answer is wrong, can someone please tell me what I am doing wrong. Thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Oct 9, 2009 #2

    Andrew Mason

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    You have to resolve all the forces into components parallel to and perpendicular to the inclined plane surface.

    The normal force is not the y component of the gravitational force. That is where you are going wrong.

    AM
     
  4. Oct 9, 2009 #3
    So Fn= Fgy-Fy
     
  5. Oct 9, 2009 #4
    Can anybody else shed some light please?
     
  6. Oct 9, 2009 #5

    Andrew Mason

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    No. The normal force is equal and opposite to the sum of all the forces acting in a perpendicular direction toward the inclined plane surface. Those forces are the sum of the components of the applied force and gravity acting in that direction.

    To calculate these forces you must draw a diagram showing the applied force, friction, normal and gravitational forces and resolve each of these into components perpendicular and parallel to the inclined plane surface.

    For example, the component of gravity acting toward the inclined plane surface is [itex]mg\cos{\theta}[/itex] and the component acting parallel to the surface is [itex]mg\sin{\theta}[/itex] where [itex]\theta[/itex] is the angle of inclination of the plane above the horizontal.

    Work out these components for the applied force to then calculate the normal force.

    The other component of the applied force pushes up the inclined plane. Keep in mind that the components of gravity and static friction forces parallel to the surface balance it.

    AM
     
    Last edited: Oct 9, 2009
  7. Oct 9, 2009 #6
    Yeah, but what other forces are acting in a perpendicular direction toward the inclined plane surface besides gravity and the horizontal force F?
     
  8. Oct 9, 2009 #7
    So Fy = Fn - Fgy
     
  9. Oct 9, 2009 #8
    I think I got it:
    The x component:
    Fx= Mu(mgCosTheta + Fy) - mgSinTheta

    The Y component:
    Fy = mgCosTheta - mgSinTheta
     
  10. Oct 9, 2009 #9
    wait no the Y component is:

    Fy = Fn -mgSinTheta
     
  11. Oct 9, 2009 #10
    Please somebody, I truly don't understand this problem.
     
  12. Oct 10, 2009 #11

    Andrew Mason

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    Why are you still trying to find the y and x components? You have to find the components parallel to and perpendicular to the inclined plane surface.

    Have you drawn a diagram showing all the forces?

    What forces are there acting on this block? What do they sum to?
    (Hint: gravity, normal, applied, static friction)

    Have you resolved all the forces into components perpendicular and parallel to the surface?

    What are the force components parallel to the surface? What do they sum to? (Hint: (Caution: the static friction force is the Maximum static friction force that the surface can provide.)

    What are the force components perpendicular to the surface? What do they sum to?

    If you do all the above and answer the questions you will solve the problem.

    AM
     
  13. Oct 10, 2009 #12

    Andrew Mason

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    What are the other forces? Do any of them act toward the surface?

    What is the component of gravity perpendicular to the surface (I gave you the answer above)? What is the component of gravity along the surface? (I gave it to you above). What is the component of the applied force perpendicular to the surface. parallel? What do they sum to? (hint: think of the normal force balancing these forces).

    AM
     
  14. Oct 10, 2009 #13
    Thanks, but my teacher said when drawing the free body diagrams for inclines, make the x-axis the inclined surface and draw everything according to that because it will be easier for us. Also, by the applied force, do you mean the horizontal force F that keeps the block in equilibrium.
     
  15. Oct 10, 2009 #14

    Andrew Mason

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    Ok. That is fine. So long as we understand that x is parallel to the plane surface and y is perpendicular.

    Yes, applied force is the horizontal force that is applied to the block. What is the component in the x direction - ie parallel to the surface?

    AM
     
  16. Oct 10, 2009 #15
    The component in the x direction is:
    Fgx - Ffr - Fx = mgSinTheta - muFn -Fx
     
  17. Oct 11, 2009 #16

    Andrew Mason

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    Ok. What is Fx in terms of the applied horizontal force, F and theta?

    What is the normal force Fn in terms of Fg and F and theta?

    AM
     
  18. Oct 11, 2009 #17
    Fx= F(CosTheta)

    Fn= Fgy +Fy
    Fn= mgCosTheta + F(SinTheta)
     
  19. Oct 12, 2009 #18

    Andrew Mason

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    Ok. What is the equation that relates all the forces acting parallel to the surface? (along the x axis).

    Ok. So you can now work out the maximum static friction.

    Plug that value into the equation you found relating all the forces acting parallel to the surface and solve for F.

    AM
     
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